Three circles are drawn, so that each circle is externally tangent to the other two circles. Each circle has a radius of $2.$ A triangle is then constructed, such that each side of the triangle is tangent to two circles, as shown below. Find the perimeter of the triangle. [asy] unitsize(1 cm); draw(Circle(dir(90),sqrt(3)/2)); draw(Circle(dir(90 + 120),sqrt(3)/2)); draw(Circle(dir(90 + 240),sqrt(3)/2)); draw((1 + sqrt(3))*dir(90)--(1 + sqrt(3))*dir(90 + 120)--(1 + sqrt(3))*dir(90 + 240)--cycle); [/asy]

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

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i am unable to solve this problem

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you probably already have the answer but anyways it's b. 15.0

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no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.